The roots of the right-hand side of an autonomous equation are the states in which nothing changes. They are solutions — the simplest ones the equation has — and they are also the scaffolding for everything else, because no other solution can cross them. Finding them is solving one algebraic equation, and once they are found the qualitative picture is nearly complete.

A ball on a hilly landscape is at rest at the bottom of a valley and at the top of a ridge. Both are equilibria; only one of them survives a nudge. The algebra finds both at once and says nothing yet about which is which, and that is the next lesson.

The field of $y'=\tfrac45y(2-y)$. The marks are flat exactly along the two levels where the right-hand side vanishes, and those two horizontal lines are themselves solutions.

An equilibrium of $y'=f(y)$ is a number $c$ with $f(c)=0$. The constant function is then a solution, called an equilibrium solution; in the plane it is a horizontal line, and where the uniqueness condition holds no other solution meets it. The equilibria are found by solving one algebraic equation and nothing else.

• An equilibrium is a value of the unknown, not of the input. Asking where a solution is in equilibrium is asking at what height, never at what time.
• $y'=y^{2}-4$ has equilibria at $2$ and $$. has none, because the exponential is never zero, and every solution of it is strictly increasing.

An equilibrium is a height at which the rate is zero. Finding them is solving one algebraic equation in the unknown.

If f had opposite signs at two heights between neighbouring roots, it would vanish between them, and that would be another root. So on each strip the motion is all one way.

Great work! You now know how to:

• find every equilibrium by solving one algebraic equation
• say why an equilibrium is a height and never a time
• use the sign of the right-hand side on a strip to say which way solutions move
• explain why an equilibrium level cannot be crossed